Problem: If you flip a fair coin $7$ times, what is the probability that you will get exactly $4$ tails?
One way to solve this problem is to figure out how many ways you can get exactly $4$ tails, then divide this by the total number of outcomes you could have gotten. Since every outcome has equal probability, this will be the probability that you will get exactly $4$ tails. How many outcomes are there where you get exactly $4$ tails? Try thinking of each outcome as a $7$ -letter word, where the first letter is "H" if the first coin toss was heads and "T" if it was tails, and so on. So, the number of outcomes with exactly $4$ tails is the same as the number of these words which have $4$ T's and $3$ H's. How many of these are there? If we treat all the letters as unique, we'll find that there are $7!$ different arrangements, overcounting $4!$ times for every time we only switch the T's around, and $3!$ times for every time we only switch the H's around. [ Show me why Let's say we toss a coin 5 times, and get tails three times. How many different re-arrangements are there of the letters "HHTTT"? Well, we have five choices for the first slot, four for the second slot, and so on, resulting in $5\cdot4\cdot3\cdot2\cdot1 = 5! = 120 \;$ different re-arrangements. But, this treats all the letters as unique, when is the same as , and and so on. So really, we need to replace all these different re-arrangements where we only move the tails around with one re-arrangement, HTHTT. There are $3!$ of these multi- colored arrangements for every normal one, so that means dividing our first guess of $5!$ by $3!$ . By the exact same logic, we need to divide by $2!$ to avoid overcounting every permutation where we just move the heads around. So, the number of re-arrangements is $\dfrac{5!}{3!2!} = \binom{5}{3}$ So, there are $\dfrac{7!}{4!3!} = 35$ different outcomes where you get exactly $4$ tails. [ How many total outcomes are there? Well, if you only flip one coin, there are two outcomes, if you flip two there are four outcomes, if you flip three there are eight. Each time you flip another coin, you double the number of possible outcomes. Altogether, there are $2^{7} = 128$ total possible outcomes. So, the probability that you will get exactly 4 tails is $\dfrac{35}{128}$.